question_answer

An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 10) From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length of the belt that is still in contact with the pulley. Also find the shaded area. (use \[\pi =3.14\] and  \[\sqrt{3}=1.73\])

Answer:

Given, a circular pulley of radius 5 cm with centre O.

\[\therefore AO=OB=OC=5\text{ }cm\]

and                   \[OP=10\text{ }cm\]

Now, in right \[\Delta \text{ }AOP\]

\[\cos \,i=\frac{AO}{OP}=\frac{5}{10}=\frac{1}{2}\]

\[\therefore i={{\cos }^{-1}}\left( \frac{1}{2} \right)=60{}^\circ \]

\[\therefore \,\,\angle AOB=2i=120{}^\circ \]

\[\Rightarrow \]Reflex \[\angle AOB=360{}^\circ -120{}^\circ =240{}^\circ \]

Length of major arc \[\overset\frown{AB}=\frac{2\pi r}{360{}^\circ }\text{reflex}\,\angle AOB\]

\[=\frac{2\times 3.14\times 5\times 240{}^\circ }{360{}^\circ }\]

\[=20.93\,\,cm\]

Hence, length of the belt that is still in contact with pulley \[=20.93\text{ }cm\]

Now, by pythagorus theorem

\[{{(AP)}^{2}}={{(OP)}^{2}}-{{(AO)}^{2}}\]

\[{{(AP)}^{2}}={{(10)}^{2}}-{{(5)}^{2}}\]

\[AP=\sqrt{100-25}\]

\[=\sqrt{75}=5\sqrt{3}\,cm\]

\[\therefore \]      \[\text{Area}\,\,\text{of}\,\,\Delta AOP=\frac{1}{2}\times 5\times 5\sqrt{3}\]

\[=\frac{25\sqrt{3}}{2}c{{m}^{2}}\]

Also,     Area of \[\text{Area}\,\,\text{of}\,\,\Delta \,BOP=\text{Area}\,\,\text{of}\,\,\Delta \,AOP\]

And,     Area of quad. \[\text{AOBP=2(Area}\,\,\text{of}\,\,\Delta \,\text{AOP)}\]

\[=2\times \frac{25\sqrt{3}}{2}=25\sqrt{3}\,\,c{{m}^{2}}\]

\[=43.25\,\,c{{m}^{2}}\]

Area of sector \[ACBO=\frac{\pi {{r}^{2}}\angle AOB}{360{}^\circ }\]

\[=\frac{3.14\times 5\times 5\times 120}{360{}^\circ }\]

\[=26.16\,c{{m}^{2}}\]

\[\therefore \] Area of shaded region = Area of quad. \[AOBP\]\[-\]Area of sector \[ACBO\]

\[=(43.25-26.16)\,c{{m}^{2}}\]

\[=17.09\,c{{m}^{2}}\]